To prepare 1.00 L of 14g BaCl2, what Molarity would it be?

The molarity would be 0,067M.

Explanation:

The molarity corresponds to a concentration measurement that expresses the moles of solute in 1 liter of solution. To calculate the molarity we must first calculate the weight of 1 mole of BaCl2, obtaining the atomic weights of Ba and Cl from the periodic table:

Weight 1 mol BaCl2= Weight Ba + 2 x(Weight Cl)= 137, 33 g + 2 x35, 45 g=

Weight 1 mol BaCl2= 208, 23 g/mol

208, 23 g1 mol BaCl2

14 g         x= (14 g   x 1 mol BaCl2)/208, 23 g=0,067 mol NaCl2

We have 0,0632 mol of BaCl2 in 1,00 Liter of solution; thus the solution is 0,067 M.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.